A tuning fork is vibrating at 250 Hz. The length of the shortest closed organ pipe that will resonate with the tuning fork will be
(Take speed of sound in air as `340 ms^( -1)`)

Options

A. 31 cm
B. 32 cm
C. 33 cm
D. 34 cm

Answer

To solve the problem of finding the length of the shortest closed organ pipe that will resonate with a tuning fork vibrating at 250 Hz, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Frequency of the tuning fork, \( f = 250 \, \text{Hz} \)
- Speed of sound in air, \( v = 340 \, \text{m/s} \)

2. **Understand the Resonance Condition for a Closed Organ Pipe:**
- The fundamental frequency (first harmonic) of a closed organ pipe is given by the formula:
\[ f = \frac{(2n + 1)v}{4L} \]
where \( n \) is the harmonic number (for the fundamental frequency, \( n = 0 \)).

3. **Substituting for the Fundamental Frequency:**
- For the fundamental frequency (first harmonic), we set \( n = 0 \):
\[ f = \frac{1 \cdot v}{4L} \]
- Rearranging this formula to find \( L \):
\[ L = \frac{v}{4f} \]

4. **Plugging in the Values:**
- Substitute \( v = 340 \, \text{m/s} \) and \( f = 250 \, \text{Hz} \):
\[ L = \frac{340}{4 \times 250} \]

5. **Calculating the Length:**
- Calculate \( 4 \times 250 = 1000 \):
\[ L = \frac{340}{1000} = 0.34 \, \text{m} \]

6. **Convert the Length to Centimeters:**
- Since \( 1 \, \text{m} = 100 \, \text{cm} \):
\[ L = 0.34 \, \text{m} \times 100 = 34 \, \text{cm} \]

### Final Answer:
The length of the shortest closed organ pipe that will resonate with the tuning fork is **34 cm**.

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